You mention you're "okay with" (a) and (b):
(a) Just to double-check, you should have seen that it is not valid, as it asserts the converse of the first statement, $q\rightarrow p$, which is not equivalent to the first statement of the form $p \rightarrow q$.
- The inverse of $\;p\rightarrow q\;$ is $\;\lnot p \rightarrow \lnot q$, which is equivalent to (is the contrapositive of), the the converse of the implication. Neither the converse of an implication nor its inverse are equivalent to the original implication.
(b) is valid, as it states the equivalent in the form of the contrapositive of the first statement. That is, if we are given $p \rightarrow q$, then it is validly asserting $\lnot q \rightarrow \lnot p$.
c) If $n$ is a real number with $n>2$, then $n^2>4$. Suppose that $n≤2$. Then $n^2≤4$.
Let "$n$ is a real number with $n > 2$" be denoted by $p$.
Let "$n^2>4$" be denoted by $q$.
Then $n \leq 4 \equiv \lnot p$. The negation of "$n> 2$" is " n is not greater than 2, i.e., $n \leq 2$.
And $n^2 \leq 4 \equiv \lnot q$. Can you see why? The negation of "$n^2 > 4$ is "$n^2$ is not greater than 4", i.e., $n^2 \leq 4$.
The first statement in (c) is of the form $p\rightarrow q$. Then we have the assertion $\lnot p$, and the conclusion therefore $ \lnot q$.
$$[(p \rightarrow q) \land \lnot p] \not \rightarrow \lnot q$$
Hence, this represents the fallacy of denying the hypothesis:
- $p\rightarrow q$
- $\lnot p$
- $\therefore \lnot q$
which is not a valid inference.
The contrapositive of the first statement in (c) would be "If $n^2 \leq 4$, then $n \leq 2$, that is, if would be of the form $\lnot q \rightarrow \lnot p$.